1. CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC

2. K-Map (1)  Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.  This will replace Boolean reduction when the circuit is large.  Write the Boolean equation in a SOP form first and then place each term on a map.

3. K-Map (2)  A graphical tool for simplifying Boolean expression.  Divided into cells, each representing a specific combination of variable values.  Number of cells = total possible combinations of the variables in an expression  Eg: 3 variables = 2 3 = 8 cells 4 variables = 2 4 = 16 cells

4.  The map is made up of a table of every possible SOP using the number of variables that are being used.  If 2 variables are used then a 2X2 map is used  If 3 variables are used then a 4X2 map is used  If 4 variables are used then a 4X4 map is used  If 5 Variables are used then a 8X4 map is used K-Map (3)

5. K-Map SOP Minimization

6. Step 1 : Place a 1 in each cell corresponding to a minterm in the SOP expression. Step 2 : Group all 1s that are adjacent cells into groupings of the maximum number of cells. A group can only contain a number of 1s equal to a power of two or groups can overlap. Step 3 : For each group, determine the variables that are the same. A complemented and an uncomplemented variable are not the same and cancel each other. Step 4 : Write the resulting product terms for each group, using only the variables that are the same within the group. Step 5 : Combine the resulting product terms into an SOP expression which will be equivalent to the original SOP expression. Simplifying an SOP Expression

7. AAAA B Notice that the map is going false to true, left to right and top to bottom The upper right hand cell is A B if X= A B then put an X in that cell AAAA B 1 This show the expression true when A = 0 and B = 0 0 1 2 3 2 Variables K-Map (1)

8. If X=AB + AB then put an X in both of these cells AAAA B 1 1 From Boolean reduction we know that A B + A B = B From the Karnaugh map we can circle adjacent cell and find that X = B AAAA B 1 1 2 Variables K-Map (2)

9. 3 Variables K-Map (1) Gray Code 00A B 01A B 11 A B 10A B 0 1 C C 0 1 0 1 2 3 2 3 6 7 6 7 4 5 4 5

10. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Each 3 variable term is one cell on a 4 X 2 Karnaugh map 11 11 3 Variables K-Map (2)

11. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C One simplification could be X = A B + A B 11 1 1 3 Variables K-Map (3)

12. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around 11 11 3 Variables K-Map (4)

13. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C The Best simplification would be X = B 11 11 3 Variables K-Map (5)

14. 3 Variables K-Map (6)  Use a K-Map to simplify the following 3- variable minterm SOP expression: X = A B C + A B C + A B C + A B C + A B C X = B + AC  Answer: X = B + AC

15. On a 3 Variables K-Map  One cell requires 3 Variables  Two adjacent cells require 2 variables  Four adjacent cells require 1 variable  Eight adjacent cells is a 1

16. 4 Variables K-Map Gray Code 00A B 01A B 11A B 10A B 0 0 0 1 1 1 1 0 C D C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10

17. 4 Variables K-Map Gray Code 00A B 01A B 11A B 10A B 0 0 0 1 1 1 1 0 C D C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10 B B A A C C D D D

18. Gray Code 00A B 01A B 11A B 10A B 0 0 0 1 1 1 1 0 C D C D 1 1 1 1 1 1 X = ABD + ABC + CD Now try it with Boolean reductions Simplify: Simplify: X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D

19. On a 4 Variables K-Map  One Cell requires 4 variables  Two adjacent cells require 3 variables  Four adjacent cells require 2 variables  Eight adjacent cells require 1 variable  Sixteen adjacent cells give a 1 or true

20. Simplify Simplify : Z = B C D + B C D + C D + B C D + A B C Gray Code 00A B 01A B 11A B 10A B 00 01 11 10 C D C D 1111 11 11 11 1 1 Z = C + A B + B D

21. Simplify using K-Map (1) Firstly, change the circuit to an SOP expression

22. Y= A + B + B C + ( A + B ) ( C + D) Y = A B + B C + A B ( C + D ) Y = A B + B C + A B C + A B D Y = A B + B C + A B C A B D Y = A B + B C + (A + B + C ) ( A + B + D) Y = A B + B C + A + A B + A D + B + B D + AC + C D Simplify using K-Map (2) SOP expression How about standard SOP expression???

23. Gray Code 00A B 01A B 11A B 10A B 00 01 11 10 C D C D 1 1 1 1 11 1 1 Y = 1 1 11 1 1 1 1 1 Simplify using K-Map (3)

24. K-Map POS Minimization

25. 3 Variables K-Map (1) Gray Code 0 0 1 1 1 0 0 1 AB C 0 1 0 1 2 3 2 3 6 7 6 7 4 5 4 5

26. 3 Variables K-Map (2)

27. 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10 4 Variables K-Map (1)

28. 4 Variables K-Map (2)

29. 4 Variables K-Map (3)

30. Mapping a Standard SOP expression  Example : Answer : Mapping a Standard POS expression  Example : Using K-Map, convert the following standard POS expression into a minimum SOP expression Answer: Y = AB + AC or standard SOP: K-Map - Examples

31. K-Map with “Don’t Care” Conditions (1) Input Output Example : 3 variables with output “don’t care (X)”

32. 4 variables with output “don’t care (X)” K-Map with “Don’t Care” Conditions (2)

33. “Don’t Care” Conditions  “Don’t Care” Conditions  Example: Determine the minimal SOP using K-Map: Answer: K-Map with “Don’t Care” Conditions (3)

34. Solution : AB CD 00 01 11 10 00 01 11 10 0 1 1 0 1 X 1 0 X X 0 0 1 0 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 Minimum SOP expression is CD AD BC

35. Exercise  Minimize this expression with a K-Map ABCD + ACD + BCD + ABCD

36. K-map POS & SOP Simplification Example: Simplify the Boolean function F (ABCD) =  (0,1,2,5,8,9,10) in (a) S-of-P (b) P-of-S Using the minterms (1’s) F (ABCD) = B’D’+B’C’+A’C’D Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De- Morgen’s theorem to get F. F’ (ABCD) = BD’+CD+AB F (ABCD) = (B’+D)(C’+D’)(A’+B’)

37. 5 variable K-map (1)  5 variables -> 32 minterms, hence 32 squares required

38. 5 variable K-map (2) Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11. The centre line must be considered as the centre of a book, each half of the K-map being a page The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

39. 5 variable K-Map (3) Example: Simplify the Boolean function F (ABCDE) =  (0,2,4,6,11,13,15,17,21,25,27,29,31) Soln: F (ABCDE) = BE+AD’E+A’B’E’

40. 6 variable K-map  6 variables -> 64 minterms, hence 64 squares required